How to find the sum of $\frac{1}{(1)(3)} + \frac{2}{(1)(3)(5)} +\frac{3}{(1)(3)(5)(7)} +\cdots$ up to n terms

Question

How do I find the sum upto n terms of the following series $$\frac{1}{(1)(3)} + \frac{2}{(1)(3)(5)} +\frac{3}{(1)(3)(5)(7)} +\cdots$$

I found the nth term to be $\frac{n}{(1)(3)(5)\cdots(2n+1)}$

Answer 1

Hint:

Like Calculate$ \frac{17}{75×76}+\frac {17×18}{75×76×77}+\frac {17×18×19}{75×76×77×78}+...$

let $$\dfrac n{1\cdot3\cdot5\cdot(2n-1)(2n+1)}=f(n)-f(n+1)$$

where $f(m)=\dfrac a{1\cdot3\cdot5\cdot(2m-1)}$

Using Telescoping series $$\sum_{r=1}^n\dfrac r{1\cdot3\cdot5\cdot(2r-1)(2r+1)}=f(1)-f(n+1)$$

Again, $$f(n)-f(n+1)=\dfrac{a(2n+1)-a}{1\cdot3\cdot5\cdot(2n-1)(2n+1)}$$

We need $2a=1$